jQuery .css() fallbacks?

Question

What is the proper way to do fallbacks with .css()? This would most commonly be used for font fallbacks but I am using it for cursor images.

Here's what I got that isn't working:

$("#dragable").css('cursor','url(images/grabbing.gif), -moz-grabbing, auto');

**UPDATE: can someone tell me the valid CSS for starters?

What I have:

cursor: url(images/grabbing.gif), -moz-grabbing, auto;

... doesn't work.**

Answer 1

Using dynamic inline styles not always a good idea. Consider dynamic classes. Define somewhere in css specific class for draggable element:

.cur-draggable { cursor: url(images/grabbing.gif), -moz-grabbing, auto; }

And try to add this class, instead of style itself;

$("#dragable").addClass('cur-draggable');

Answer 2

Not sure why that breaks it, but the values aren't being written to the DOM.

Though, the css() function is really just a shortcut to editing the "style" attribute inline on the element, so you could do this to get the same result:

$("#dragable").attr('style','cursor: url(images/grabbing.gif), -moz-grabbing, auto');